线段树练习：【SPOJ1716】 GSS3

原题：http://www.spoj.com/problems/GSS3/

SPOJ Problem Set (classical) 1716. Can you answer these queries III Problem code: GSS3

You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations:

modify the i-th element in the sequence or for given x y print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

Input

The first line of input contains an integer N. The following line contains N integers, representing the sequence A1..AN.

The third line contains an integer M. The next M lines contain the operations in following form:

0 x y: modify Ax into y (|y|<=10000).

1 x y: print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

Output

For each query, print an integer as the problem required.

Example

Input:
4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3

Output:
6
4
-3

Added by:	Bin Jin
Date:	2007-08-03
Time limit:	0.330s
Source limit:	5000B
Memory limit:	1536MB
Cluster:	Cube (Intel Pentium G860 3GHz)
Languages:	All except: C++ 4.9 SCM chicken
Resource:	own problem

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int MAXD = 10010, INF = 0x3f3f3f3f;

struct Seg
{
int mid, l, r;
};

Seg seg[4 * MAXD];

int A[MAXD], a[MAXD];

void update(int cur, int x, int y)
{
int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
seg[cur].mid = max(seg[ls].mid, seg[rs].mid);
seg[cur].mid = max(seg[cur].mid, seg[ls].r + seg[rs].l);
seg[cur].l = max(seg[ls].l, A[mid] - A[x - 1] + seg[rs].l);
seg[cur].r = max(seg[rs].r, A[y] - A[mid] + seg[ls].r);
}

void build(int cur, int x, int y)
{
int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
if (x == y)
{
seg[cur].mid = seg[cur].l = seg[cur].r = a[x];
return ;
}
build(ls, x, mid);
build(rs, mid + 1, y);
update(cur, x, y);
}

int query(int cur, int x, int y, int s, int t, int flag, int &ans)
{
int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
if (x >= s && y <= t)
{
ans = max(ans, seg[cur].mid);
if (flag == 0) return seg[cur].l;
else return seg[cur].r;
}
if (mid >= t) return query(ls, x, mid, s, t, 0, ans);
else if (mid + 1 <= s) return query(rs, mid + 1, y, s, t, 1, ans);

int ln, rn;
ln = query(ls, x, mid, s, t, 1, ans);
rn = query(rs, mid + 1, y, s, t, 0, ans);
ans = max(ans, ln + rn);

if (flag == 0) return max(seg[ls].l, A[mid] - A[x - 1] + rn);
else return max(seg[rs].r, A[y] - A[mid] + ln);
}

void Search(int cur, int x, int y, int s, int t, int flag, int &ans)
{
int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
if (x >= s && y <= t)
{
if (flag == 0) ans = max(ans, A[x - 1] - A[s - 1] + seg[cur].l);
else ans = max(ans, A[t] - A[y] + seg[cur].r);
return;
}
if (mid >= s) Search(ls, x, mid, s, t, flag, ans);
if (mid + 1 <= t) Search(rs, mid + 1, y, s, t, flag, ans);
}

int main(void)
{
int N, q, ans, t1, t2, x1, y1, x2, y2;

scanf("%d", &N);
A[0] = 0;
for (int i = 1; i <= N; i ++)
{
scanf("%d", &a[i]);
A[i] = A[i - 1] + a[i];
}
build(1, 1, N);

scanf("%d", &q);
for (int i = 0; i < q; i ++)
{
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (y1 < x2)
{
t1 = t2 = -INF;
Search(1, 1, N, x1, y1, 1, t1);
Search(1, 1, N, x2, y2, 0, t2);
printf("%d\n", t1 + t2 + A[x2 - 1] - A[y1]);
}
else
{
ans = -INF;
query(1, 1, N, x2, y1, 0, ans);
t1 = t2 = -INF;

Search(1, 1, N, x1, y1, 1, t1);
Search(1, 1, N, y1, y2, 0, t2);

ans = max(ans, t1 + t2 - a[y1]);
t1 = t2 = -INF;

Search(1, 1, N, x1, x2, 1, t1);
Search(1, 1, N, x2, y2, 0, t2);

ans = max(ans, t1 + t2 - a[x2]);
printf("%d\n", ans);
}
}

return 0;
}

与GSS1类似，单点更新直到叶子，然后一层层PushUp